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Prove the identity

  1. Dec 12, 2011 #1
    1. The problem statement, all variables and given/known data

    sin2x-tan2x=-2(sinx)^2(tan2x)

    2. The attempt at a solution

    I have like two pages of attempts, but I don't know if it would be useful to copy it into the forum. :|
     
  2. jcsd
  3. Dec 12, 2011 #2
    [tex]\tan 2x = \frac{\sin 2x}{\cos 2x}[/tex]
    [tex]\cos 2x = \cos^2x-\sin^2x[/tex]
    [tex]1=\sin^2x+\cos^2x[/tex]
    That is all you need to know.
     
  4. Dec 12, 2011 #3

    eumyang

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    Homework Helper

    Hint #1: I would start on the left side and rewrite tan 2x in terms of sine and cosine.
    Hint #2: Near the end, you should have tan 2x (cos 2x - 1). Look at the formula for cos 2x and you can see cos 2x - 1 will equal to.
     
  5. Dec 12, 2011 #4
    Okay, I got to tan2x(cos2x -1). :)

    I sub'd cos2x -1 = -(sinx)^2

    And I'm left with tan2x - (sinx)^2.

    edit: oh no wait! I got it! Thanks!
     
  6. Dec 12, 2011 #5
    I hate identities. They make me feel so stupid. How did you guys do it so fast? How did you know you had to start on the left side? Trig. identities feels like just a bunch of trial & error (I hate that).

    edit: By the way, I like how you guys didn't just give me the answer. It feels to good to actually finish do an identity problem for once. :D
     
  7. Dec 12, 2011 #6
    For homework: Solve [tex]\tan x + \tan 2x + \tan 3x = 0[/tex] :)
     
  8. Dec 12, 2011 #7
    Hmm... I'm trying it, and it seems really complicated. Can you just tell me if I'm on the right track by doing:

    [itex]tan2x=\frac { 2tanx }{ 1-{ tan }^{ 2 }x } [/itex]

    and

    [itex] tan2x=\frac { \frac { 2tanx }{ 1-{ tan }^{ 2 }x } +tanx }{ 1-\frac { 2{ tan }^{ 2 }x }{ 1-{ tan }^{ 2 }x } } [/itex]
     
    Last edited: Dec 12, 2011
  9. Dec 12, 2011 #8
    It seems well.
    Also, you can solve using: [tex]\frac{\sin x}{\cos x} + \frac{\sin {2x}}{\cos {2x}} = -\frac{\sin {3x}}{\cos {3x}}[/tex] (but, this is harder way I think)
     
  10. Dec 12, 2011 #9

    eumyang

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    Homework Helper

    Proving trig identities does take trial & error a lot of the time. For me it took a lot of practice to get the hang of it. You'll start recognizing the different variants of the identities, like the one I suggested for cos 2x (cos2x -1 = -sin2 x). Now that I teach precalculus at a high school, I subject my own students every year to the same torture of proving identities. :rofl:
     
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